If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra 2>Unit 1

Lesson 2: Average rate of change of polynomials

# Sign of average rate of change of polynomials

Discover how to find the average rate of change in polynomials. Learn to identify intervals with positive average rates of change by comparing function values at different points. See how this concept applies to real-life situations, making math fun and practical!

## Want to join the conversation?

• I'm confused what it means when it Sal says that it should be `h(xf) > h(xi)` when all the answers are in a `a ≤ x ≤ b` format.

Also I'm not sure if Sal went over this in the video, but does having a negative number in `a` in `a ≤ x ≤ b` spot and a 0 or higher in the `b` spot mean it will be positive?
• The `h(xf) > h(xi)` is referring to `a ≤ x ≤ b`, where `h(xf)` is `b` and `h(xi)` is a.

Basically it is saying that in this situation for an answer to be correct, the output of `h(b)` has to be bigger than `h(a)` to warrant a positive change. If you were looking for a negative change it would be the other way around.

To answer your second question that might not always be the case.

Say you have `f(x) = -1x^2`

If the interval is `-1 ≤ x ≤ 1` then:

`f(-1) = -1` and `f(1) = -1`, making it not positive
• I don't understand any of this. I've watched the video numerous times and it still doesn't make sense. Can someone please help me?
• for the first What he is saying is that H is our function and what we do is multiply by our integer, which is 0, so we substitute x for zero, so the start function is = to zero. He does the same thing for the next integer, which is 2 and gets -3. That is a negative decline, so the interval has a negative average rate.
• hi, how is this relevant
• only god knows
• my head is exploding am not understanding anything
• don't worry bro I'm goin straight from algebra to algebra two and I can barely grasp this concept
• you lost me 44 secs into this video
• I'm not understanding the sign of interval change problem
problem A.) for example

h(x)=1/8 x^3-x^2
how did h(0)=0, h(2)=-3
• h(x) takes the place of y on an x-y plane, so when we're taking h(0), we're looking at the line the function makes and seeing what the number on the vertical axis is when the number on the horizontal axis is 0. In the equation, we're seeing what y (which is h(x)) is when x is 0. So we replace all the x's in the equation with 0, and then solve it. 0 · 1/8 is 0, and 0 to any exponent is also 0. So it ends up as 0 · 0, which is 0.
• Just wondering... is there a faster way to do this than to go over all of the multiple choice options?
• Not really, it's kind of like guess and check in a way you have to go through each one individually
• How do you know which x value is the x final, and which x value is the x initial?